What is #cos(arccos (3/5) - arcsin (4/5))#?

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Answer 1 · 2018-07-14T13:36:03

# 1.#

Let, #arcsin(4/5)=alpha#.

#:. sinalpha=4/5, where, alpha in [-pi/2,pi/2]=Q_1uuQ_4#.

But, since #sinalpha=4/5 gt 0, alpha !in Q_4 :. alpha in Q_1#.

Now, #cos^2alpha=1-sin^2alpha=1-(4/5)^2=9/25#.

Knowing that, #alpha in Q_1, cosalpha=+sqrt(9/25)=+3/5#.

Thus, #cosalpha=3/5, where, alpha in Q_1=[0,pi/2] sub [0,pi]#.

#:." by definition, "arccos(3/5)=alpha=arcsin(4/5)#.

# rArr (arccos(3/5)-arcsin(4/5))=0#.

Consequently, #cos(arccos(3/5)-arcsin(4/5))=cos0=1#.

#color(purple)("Enjoy Maths.!")#

Answer 2 · 2018-07-14T15:19:24

#1#

#cos((arccos)(3/5)-arc sine (4/5))#

#:.=cos(#(arccos(0.6)-arc sine (0.8))#

#:.=cos(53^@7'48''-53^@7'48'')#

#:.cos(0)=1#